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What is the rectangular form of the polar equation?

r−rcosθ=6

 Apr 30, 2020
 #1
avatar
0

\(\frac{1}{12}{y}^{2}-3\)

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 Apr 30, 2020
 #2
avatar+111435 
+2

r + r cosθ = 6

 

r = √[x^2 + y^2]

 

cos θ =  x/r 

 

So we have that

 

√[x^2 + y^2]  +  r (x/r)  =  6

 

√[x^2 + y^2] + x  = 6

 

√[x^2 + y^2]  =  6 - x        square both sides

 

x^2  + y^2  = x^2 - 12x + 36

 

y^2  = -12x + 36

 

12x = -y^2 + 36

 

x = -y^2/12 + 3

 

cool cool cool

 Apr 30, 2020
 #3
avatar+110107 
+1

Thanks Chris, I was wondering how to do that.  cool

 Apr 30, 2020
edited by Melody  Apr 30, 2020
 #4
avatar+111435 
0

LOL!!!!....I'm still wondering myself    !!!!

 

cool cool cool

CPhill  Apr 30, 2020
 #5
avatar+110107 
0

What are you wondering about?

Melody  Apr 30, 2020

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